Sachin throws a ball vertically upwards. The ball returns to sachin after 6 sec. Can you find :
1) the velocity with which sachin throws the ball
2) the maximum height it reaches
3) its position after 4 sec

(1)Time of flight, T=2ugVelocity of projection:u=gT2  =9.8×62   =29.4 m/s(2)Maximum height:H=u22g   =29.422×9.8   =44.1 m(3)Position after 4 sec:S=ut-12gt2   =29.4×4-12×9.8×42   =39.2 m from the ground

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a] - 60ms-1
b] 180m 
c] 100 m 
i think this is the answer
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Please explain the way u took out the solutions.....
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